Question

prove that product of three consecutive positive integer is divisible by 6​

Answer 1

Answer:

Step-by-step explanation:

Let the three consecutive positive integers be n, (n+1), (n+2)

Now to test the divisibility rule of 6, the number should be divisible by 2 and 3.

For the divisibility rule of 3 the sum of the digits should be divisible by 3

So, n+(n+1)+(n+2)

= 3n+3

= 3(n+1) which is divisible by 3

Also, among the three numbers one number would be definitely even which will be divisible by 2.

Hence it satisfies the divisibility rule of 6. So the product of three consecutive positive integers is divisible by 6.[proved]

Answer 2

$\huge \fbox \red{Solution:}$

Let n be any positive integer.

Thus, the three consecutive positive integers are n, n+1 and n+2.

We know that any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or

6q+5. (From Euclid's division lemma for b= 6)

So,

For n= 6q,

⇒ n(n+1)(n+2)= 6q(6q+1)(6q+2)

⇒ n(n+1)(n+2)= 6[q(6q+1)(6q+2)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6.

[m= q(6q+1)(6q+2)]

For n= 6q+1,

⇒ n(n+1)(n+2)= (6q+1)(6q+2)(6q+3)

⇒ n(n+1)(n+2)= 6[(6q+1)(3q+1)(2q+1)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6.

[m= (6q+1)(3q+1)(2q+1)]

For n= 6q+2,

⇒ n(n+1)(n+2)= (6q+2)(6q+3)(6q+4)

⇒ n(n+1)(n+2)= 6[(3q+1)(2q+1)(6q+4)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6.

[m= (3q+1)(2q+1)(6q+4)]

For n= 6q+3,

⇒ n(n+1)(n+2)= (6q+3)(6q+4)(6q+5)

⇒ n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6.

[m= (2q+1)(3q+2)(6q+5)]

For n= 6q+4,

⇒ n(n+1)(n+2)= (6q+4)(6q+5)(6q+6)

⇒ n(n+1)(n+2)= 6[(3q+2)(3q+1)(2q+2)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6.

[m= (3q+2)(3q+1)(2q+2)]

For n= 6q+5,

⇒ n(n+1)(n+2)= (6q+5)(6q+6)(6q+7)

⇒ n(n+1)(n+2)= 6[(6q+5)(q+1)(6q+7)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6.

[m= (6q+5)(q+1)(6q+7)]

=Hence, the product of three consecutive positive integers is divisible by 6.